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Popular Trigonometry >

(sin(x)+cos(x))^2>= 3-2tan(x)+tan^2(x)

Solution

(sin (x) + cos (x))^{2} \geq 3 - 2 tan (x) + tan^{2} (x)

Solution

πn \leq x < \frac{π}{2} + πn

Interval Notation

[πn, \frac{π}{2} + πn)

Decimal

πn \leq x < 1.57079 \dots + πn

Solution steps

(sin (x) + cos (x))^{2} \geq 3 - 2 tan (x) + tan^{2} (x)

Subtract

- 2 tan (x) + tan^{2} (x)

from both sides

(sin (x) + cos (x))^{2} - (- 2 tan (x) + tan^{2} (x)) \geq 3 - 2 tan (x) + tan^{2} (x) - (- 2 tan (x) + tan^{2} (x))

(sin (x) + cos (x))^{2} - (- 2 tan (x) + tan^{2} (x)) \geq 3

Use the following identity:

cos (x) + sin (x) = 2 sin (\frac{π}{4} + x)

(2 sin (\frac{π}{4} + x))^{2} - (tan^{2} (x) - 2 tan (x)) \geq 3

Simplify

(2 sin (\frac{π}{4} + x))^{2} - (tan^{2} (x) - 2 tan (x)) : 2 sin^{2} (\frac{π}{4} + x) - tan^{2} (x) + 2 tan (x)

(2 sin (\frac{π}{4} + x))^{2} - (tan^{2} (x) - 2 tan (x))

(2 sin (\frac{π}{4} + x))^{2} = 2 sin^{2} (\frac{π}{4} + x)

(2 sin (\frac{π}{4} + x))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= (2)^{2} sin^{2} (x + \frac{π}{4})

(2)^{2} : 2

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2

= 2 sin^{2} (\frac{π}{4} + x)

= 2 sin^{2} (x + \frac{π}{4}) - (tan^{2} (x) - 2 tan (x))

- (tan^{2} (x) - 2 tan (x)) : - tan^{2} (x) + 2 tan (x)

- (tan^{2} (x) - 2 tan (x))

Distribute parentheses

= - (tan^{2} (x)) - (- 2 tan (x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - tan^{2} (x) + 2 tan (x)

= 2 sin^{2} (\frac{π}{4} + x) - tan^{2} (x) + 2 tan (x)

2 sin^{2} (\frac{π}{4} + x) - tan^{2} (x) + 2 tan (x) \geq 3

Periodicity of

2 sin^{2} (\frac{π}{4} + x) - tan^{2} (x) + 2 tan (x) : π

The compound periodicity of the sum of periodic functions is the least common multiplier of the periods

2 sin^{2} (\frac{π}{4} + x), tan^{2} (x), 2 tan (x)

Periodicity of

2 sin^{2} (\frac{π}{4} + x) : π

Periodicity of

sin^{n} (x) = \frac{Periodicity of sin ( x )}{2},

if n is even

Periodicity of

sin (\frac{π}{4} + x) : 2 π

Periodicity of

sin (x)

2 π

= 2 π

\frac{2 π}{2}

Simplify

π

Periodicity of

tan^{2} (x) : π

P er i o d i c i t y o f tan^{n} (x) =

Periodicity of

tan (x)

Periodicity of

tan (x) : π

Periodicity of

tan (x)

π

= π

π

Periodicity of

2 tan (x) : π

Periodicity of

a \cdot tan (b x + c) + d = \frac{periodicity of tan ( x )}{∣ b ∣}

Periodicity of

tan (x)

π

= \frac{π}{∣ 1 ∣}

Simplify

= π

Combine periods:

π, π, π

= π

Express with sin, cos

2 sin^{2} (\frac{π}{4} + x) - tan^{2} (x) + 2 tan (x) \geq 3

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

2 sin^{2} (\frac{π}{4} + x) - (\frac{sin ( x )}{cos ( x )})^{2} + 2 \cdot \frac{sin ( x )}{cos ( x )} \geq 3

2 sin^{2} (\frac{π}{4} + x) - (\frac{sin ( x )}{cos ( x )})^{2} + 2 \cdot \frac{sin ( x )}{cos ( x )} \geq 3

Simplify

2 sin^{2} (\frac{π}{4} + x) - (\frac{sin ( x )}{cos ( x )})^{2} + 2 \cdot \frac{sin ( x )}{cos ( x )} : \frac{2 sin ^{2} ( \frac{π + 4 x}{4} ) cos ^{2} ( x ) - sin ^{2} ( x ) + 2 sin ( x ) cos ( x )}{cos ^{2} ( x )}

2 sin^{2} (\frac{π}{4} + x) - (\frac{sin ( x )}{cos ( x )})^{2} + 2 \cdot \frac{sin ( x )}{cos ( x )}

sin^{2} (\frac{π}{4} + x) = sin^{2} (\frac{π + 4 x}{4})

sin^{2} (\frac{π}{4} + x)

Join

\frac{π}{4} + x : \frac{π + 4 x}{4}

\frac{π}{4} + x

Convert element to fraction:

x = \frac{x 4}{4}

= \frac{π}{4} + \frac{x \cdot 4}{4}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π + x \cdot 4}{4}

= sin^{2} (\frac{π + x \cdot 4}{4})

= 2 sin^{2} (\frac{4 x + π}{4}) - (\frac{sin ( x )}{cos ( x )})^{2} + 2 \cdot \frac{sin ( x )}{cos ( x )}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= 2 sin^{2} (\frac{4 x + π}{4}) - \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 2 \cdot \frac{sin ( x )}{cos ( x )}

Multiply

2 \cdot \frac{sin ( x )}{cos ( x )} : \frac{2 sin ( x )}{cos ( x )}

2 \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( x ) \cdot 2}{cos ( x )}

= 2 sin^{2} (\frac{4 x + π}{4}) - \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + \frac{2 sin ( x )}{cos ( x )}

Convert element to fraction:

2 sin^{2} (\frac{4 x + π}{4}) = \frac{2 sin ^{2} ( \frac{4 x + π}{4} )}{1}

= \frac{2 sin ^{2} ( \frac{π + x \cdot 4}{4} )}{1} - \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ( x ) \cdot 2}{cos ( x )}

Least Common Multiplier of

1, cos^{2} (x), cos (x) : cos^{2} (x)

1, cos^{2} (x), cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear in at least one of the factored expressions

= cos^{2} (x)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos^{2} (x)

For

\frac{2 sin ^{2} ( \frac{π + x \cdot 4}{4} )}{1} :

multiply the denominator and numerator by

cos^{2} (x)

\frac{2 sin ^{2} ( \frac{π + x \cdot 4}{4} )}{1} = \frac{2 sin ^{2} ( \frac{π + x \cdot 4}{4} ) cos ^{2} ( x )}{1 \cdot cos ^{2} ( x )} = \frac{2 sin ^{2} ( \frac{π + x \cdot 4}{4} ) cos ^{2} ( x )}{cos ^{2} ( x )}

For

\frac{sin ( x ) \cdot 2}{cos ( x )} :

multiply the denominator and numerator by

cos (x)

\frac{sin ( x ) \cdot 2}{cos ( x )} = \frac{sin ( x ) \cdot 2 cos ( x )}{cos ( x ) cos ( x )} = \frac{sin ( x ) \cdot 2 cos ( x )}{cos ^{2} ( x )}

= \frac{2 sin ^{2} ( \frac{π + x \cdot 4}{4} ) cos ^{2} ( x )}{cos ^{2} ( x )} - \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ( x ) \cdot 2 cos ( x )}{cos ^{2} ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 sin ^{2} ( \frac{π + x \cdot 4}{4} ) cos ^{2} ( x ) - sin ^{2} ( x ) + sin ( x ) \cdot 2 cos ( x )}{cos ^{2} ( x )}

\frac{2 sin ^{2} ( \frac{π + 4 x}{4} ) cos ^{2} ( x ) - sin ^{2} ( x ) + 2 sin ( x ) cos ( x )}{cos ^{2} ( x )} \geq 3

Find the zeroes and undifined points of

\frac{2 sin ^{2} ( \frac{π + 4 x}{4} ) cos ^{2} ( x ) - sin ^{2} ( x ) + 2 sin ( x ) cos ( x )}{cos ^{2} ( x )}

for

0 \leq x < π

To find the zeroes, set the inequality to zero

\frac{2 sin ^{2} ( \frac{π + 4 x}{4} ) cos ^{2} ( x ) - sin ^{2} ( x ) + 2 sin ( x ) cos ( x )}{cos ^{2} ( x )} = 0

Find the undefined points:

x = \frac{π}{2}

Find the zeros of the denominator

cos^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (x) = 0

General solutions for

cos (x) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x < π

x = \frac{π}{2}

\frac{π}{2}

Identify the intervals

0 < x < \frac{π}{2}, \frac{π}{2} < x < π

Summarize in a table:

2 sin^{2} (\frac{π + 4 x}{4}) cos^{2} (x) - sin^{2} (x) + 2 sin (x) cos (x) cos^{2} (x) \frac{2 s i n ^{2} ( \frac{π + 4 x}{4} ) c o s ^{2} ( x ) - s i n ^{2} ( x ) + 2 s i n ( x ) c o s ( x )}{c o s ^{2} ( x )} x = 0 + + + 0 < x < \frac{π}{2} + + + x = \frac{π}{2} - 0 Undefined \frac{π}{2} < x < π - + - x = π + + +

Identify the intervals that satisfy the required condition:

\geq 0

x = 0 or 0 < x < \frac{π}{2} or x = π

Merge Overlapping Intervals

0 \leq x < \frac{π}{2} or x = π

The union of two intervals is the set of numbers which are in either interval

x = 0

0 < x < \frac{π}{2}

0 \leq x < \frac{π}{2}

The union of two intervals is the set of numbers which are in either interval

0 \leq x < \frac{π}{2}

x = π

0 \leq x < \frac{π}{2} or x = π

0 \leq x < \frac{π}{2} or x = π

Apply the periodicity of

2 sin^{2} (\frac{π}{4} + x) - tan^{2} (x) + 2 tan (x)

πn \leq x < \frac{π}{2} + πn

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